I'm trying to understand the proof of Rank–nullity theorem,but there are parts that I don't understand:

Steinitz exchange lemma

If ${\displaystyle U=\{u_{1},\dots ,u_{m}\}}$ is a set of ${\displaystyle m}$ linearly independent vectors in a vector space ${\displaystyle V}$, and ${\displaystyle W=\{w_{1},\dots ,w_{n}\}}$ span ${\displaystyle V}$, then:

  1. ${\displaystyle m\leq n}$;
  1. There is a set ${\displaystyle W'\subseteq W}$ with ${\displaystyle |W'|=n-m}$ such that ${\displaystyle U\cup W'}$ spans ${\displaystyle V}$.

Rank–nullity theorem

Let ${\displaystyle V}, {\displaystyle W}$ be vector spaces, where ${\displaystyle V}$ is finite dimensional. Let ${\displaystyle T\colon V\to W}$ be a linear transformation. Then

$${\displaystyle \operatorname {Rank} (T)+\operatorname {Nullity} (T)=\dim V}$$


Let ${\displaystyle V,W}$ be vector spaces over some field ${\displaystyle \mathbb {F} }$ and ${\displaystyle T}$ defined as in the statement of the theorem with ${\displaystyle \dim V=n}$.

As ${\displaystyle \operatorname {Ker} T\subset V}$ is a subspace, there exists a basis for it. Suppose ${\displaystyle \dim \operatorname {Ker} T=k}$

I know that $\text{ker T}$ is a subset of a finite set $V$,and hence is finite,but how does this imply that $\text{ker T}$ does have a basis?

and let ${\displaystyle {\mathcal {K}}:=\{v_{1},\ldots ,v_{k}\}\subset \operatorname {Ker} (T)}$ be such a basis.

We may now, by the Steinitz exchange lemma, extend ${\displaystyle {\mathcal {K}}}$ with ${\displaystyle n-k}$ linearly independent vectors ${\displaystyle w_{1},\ldots ,w_{n-k}}$ to form a full basis of ${\displaystyle V}$.


${\displaystyle {\mathcal {S}}:=\{w_{1},\ldots ,w_{n-k}\}\subset V\setminus \operatorname {Ker} (T)}$ such that

${\displaystyle {\mathcal {B}}:={\mathcal {K}}\cup {\mathcal {S}}=\{v_{1},\ldots ,v_{k},w_{1},\ldots ,w_{n-k}\}\subset V}$ is a basis for ${\displaystyle V}$.

What is $V,W,W',U$ here? Since the proof uses Steinitz exchange lemma,however I can't recognize $V,W,W',U$.

From this, we know that ${\displaystyle \operatorname {Im} T=\operatorname {Span} T({\mathcal {B}})=\operatorname {Span} \{T(v_{1}),\ldots ,T(v_{k}),T(w_{1}),\ldots ,T(w_{n-k})\}=\operatorname {Span} \{T(w_{1}),\ldots ,T(w_{n-k})\}=\operatorname {Span} T({\mathcal {S}})}$

Why $\text{Im}\; T=\text{span} \;T (\mathcal B)$?

And why $\operatorname {Span} \{T(v_{1}),\ldots ,T(v_{k}),T(w_{1}),\ldots ,T(w_{n-k})\}=\operatorname {Span} \{T(w_{1}),\ldots ,T(w_{n-k})\}$?

Thanks for your help.

  • $\begingroup$ Every finite-dimensional space has a basis. math.stackexchange.com/questions/2785769/… $\endgroup$
    – Chris Wang
    Dec 27, 2020 at 17:27
  • $\begingroup$ Second question: $V,W$ were given as your vector spaces, $W'$ is the $n-k$ linearly independent vectors $w_1,\dots,w_{n-k}$, $U$ is $\mathcal K$. Third question: Im $T$ is Span$T(\mathcal B)$ by definition; the image of a linear map is all vectors spanned by a basis of its image. Fourth question: it is because $v_1,\dots,v_k$ are in the kernel, so they don't "change" the span. $\endgroup$
    – Chris Wang
    Dec 27, 2020 at 17:32

2 Answers 2


let $v_1,v_2,...,v_n$ be a basis for $V$ then for all $v \in V$

$v=c_1v_1+...+c_nv_n$ $\Rightarrow$ $T(v)=T(C_1v_1+...+c_nv_n)=c_1T(v_1)+...+c_nT(v_n) \in \text{span} \;T (\mathcal B)$

similarly, $w \in \text{span} \;T (\mathcal B) \Rightarrow w=T(c_1v_1+...+c_nv_n) \in Im(T)$

that is why $Im(T)=\text{span} \;T (\mathcal B)$

  • $\begingroup$ yes these are the hidden steps that were not written in my proof $\endgroup$ Dec 27, 2020 at 17:52
  1. $\ker T$ is a subspace, hence it is a vector space, so it has a basis.

  2. $V$ is $V$, $W$ is any fixed basis of $V$, $U=\mathcal K$ and $W'=\mathcal S$.

  3. It's because $\mathcal B$ is a basis of $V$, thus in particular every vector is a linear combination of them, and $T$ is linear.

  4. $T(v_i)=0$ since $v_i\in\ker T$.

  • $\begingroup$ It's by construction: first let $W$ be an arbitrary basis of $V$, then consider an arbitrary basis $U$ of $\ker T$. $U$ remains linearly independent in $V$. Apply Steinitz exchange lemma to obtain $W'\subseteq W$ with the given property. $\endgroup$
    – Berci
    Dec 27, 2020 at 19:58

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